Parsing periods in a column dataframe - regex

Question

I have a csv with one of the columns that contains periods:
timespan (string): PnYnMnD, where P is a literal value that starts the expression, nY is the number of years followed by a literal Y, nM is the number of months followed by a literal M, nD is the number of days followed by a literal D, where any of these numbers and corresponding designators may be absent if they are equal to 0, and a minus sign may appear before the P to specify a negative duration.
I want to return a data frame that contains all the data in the csv with parsed timespan column.
So far I have a code that parses periods:
import re
timespan_regex = re.compile(r'P(?:(\d+)Y)?(?:(\d+)M)?(?:(\d+)D)?')
def parse_timespan(timespan):
# check if the input is a valid timespan
if not timespan or 'P' not in timespan:
return None
# check if timespan is negative and skip initial 'P' literal
curr_idx = 0
is_negative = timespan.startswith('-')
if is_negative:
curr_idx = 1
# extract years, months and days with the regex
match = timespan_regex.match(timespan[curr_idx:])
years = int(match.group(1) or 0)
months = int(match.group(2) or 0)
days = int(match.group(3) or 0)
timespan_days = years * 365 + months * 30 + days
return timespan_days if not is_negative else -timespan_days
print(parse_timespan(''))
print(parse_timespan('P2Y11M20D'))
print(parse_timespan('-P2Y11M20D'))
print(parse_timespan('P2Y'))
print(parse_timespan('P0Y'))
print(parse_timespan('P2Y4M'))
print(parse_timespan('P16D'))
Output:
None
1080
-1080
730
0
850
16
How do I apply this code to the whole csv column while running the function processing csv?
def do_process_citation_data(f_path):
global my_ocan
my_ocan = pd.read_csv(f_path, names=['oci', 'citing', 'cited', 'creation', 'timespan', 'journal_sc', 'author_sc'],
parse_dates=['creation', 'timespan'])
my_ocan = my_ocan.iloc[1:] # to remove the first row
my_ocan['creation'] = pd.to_datetime(my_ocan['creation'], format="%Y-%m-%d", yearfirst=True)
my_ocan['timespan'] = parse_timespan(my_ocan['timespan']) #I tried like this, but sure it is not working :)
return my_ocan
Thank you and have a lovely day :)

Answer 1

Like with Python's builtin map, Pandas also has that method. You can check its documentation here. Since you already have your function ready which takes a single parameter and returns a value, you just need this:
my_ocan['timespan'] = my_ocan['timespan'].map(parse_timespan) #This will take each value in the column "timespan", pass it to your function 'parse_timespan', and update the specific row with the returned value
And here is a generic demo:
import pandas as pd
def demo_func(x):
#Takes an int or string, prefixes with 'A' and returns a string.
return "A" + str(x)
df = pd.DataFrame({"Column_1": [1, 2, 3, 4], "Column_2": [10, 9, 8, 7]})
print(df)
df['Column_1'] = df['Column_1'].map(demo_func)
print("After mapping:\n{}".format(df))
Output:
Column_1 Column_2
0 1 10
1 2 9
2 3 8
3 4 7
After mapping:
Column_1 Column_2
0 A1 10
1 A2 9
2 A3 8
3 A4 7